考虑以下程序:
#include <iostream>
#include "xtensor/xarray.hpp"
#include "xtensor/xio.hpp"
#include "xtensor/xview.hpp"
xt::xarray<double> arr1
{1.0, 2.0, 3.0};
xt::xarray<double> arr2
{5.0, 6.0, 7.0};
template <typename T, typename U>
struct container{
container(const T& t, const U& u) : a(t), b(u) {}
T a;
U b;
};
template <typename T, typename U>
container<T, U> make_container(const T& t, const U& u){
return container<T,U>(t, u);
}
auto c = make_container(arr1, arr1);
std::cout << (arr1 * arr1) + arr2;
template <typename A, typename B, typename R>
auto operator+(const container<A, B>& e1, const R& e2){
return (e1.a * e1.b) + e2;
}
std::cout << (c + arr2);
如果我们看代码:
std::cout << (arr1 * arr1) + arr2;
它将输出:
{ 6., 10., 16.}
但是,运行最后一行:
std::cout << (c + arr2);
产生以下结果:
{{ 6., 9., 14.}, { 7., 10., 15.}, { 8., 11., 16.}}
为什么会这样?我将函数定义更改operator+
为以下内容:
template <typename A, typename B, typename R>
auto operator+(const container<A, B>& e1, const R& e2){
std::cout << __PRETTY_FUNCTION__ << std::endl;
return (e1.b * e1.alpha) + e2;
}
输出有点令人惊讶:
auto operator+(const container<A, B> &, const R &) [A = xt::xarray_container<xt::uvector<double, std::allocator<double> >, xt::layout_type::row_major, xt::svector<unsigned long, 4, std::allocator<unsigned long>, true>, xt::xtensor_expression_tag>, B = xt::xarray_container<xt::uvector<double, std::allocator<double> >, xt::layout_type::row_major, xt::svector<unsigned long, 4, std::allocator<unsigned long>, true>, xt::xtensor_expression_tag>, R = double]
auto operator+(const container<A, B> &, const R &) [A = xt::xarray_container<xt::uvector<double, std::allocator<double> >, xt::layout_type::row_major, xt::svector<unsigned long, 4, std::allocator<unsigned long>, true>, xt::xtensor_expression_tag>, B = xt::xarray_container<xt::uvector<double, std::allocator<double> >, xt::layout_type::row_major, xt::svector<unsigned long, 4, std::allocator<unsigned long>, true>, xt::xtensor_expression_tag>, R = double]
auto operator+(const container<A, B> &, const R &) [A = xt::xarray_container<xt::uvector<double, std::allocator<double> >, xt::layout_type::row_major, xt::svector<unsigned long, 4, std::allocator<unsigned long>, true>, xt::xtensor_expression_tag>, B = xt::xarray_container<xt::uvector<double, std::allocator<double> >, xt::layout_type::row_major, xt::svector<unsigned long, 4, std::allocator<unsigned long>, true>, xt::xtensor_expression_tag>, R = double]
{{ 6., 9., 14.}, { 7., 10., 15.}, { 8., 11., 16.}}
为什么+
在一个操作中调用了 3 个操作?是否在某处定义了导致此行为的宏?中的R
类型operator+
给了我们double
,实际上应该是xt::xarray<double>
。
任何见解将不胜感激,谢谢。